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#1
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Figure it out yourself. You have no clue how to read a polar and I am not going to teach you. Quote:
The effect of trailing edge flaps is to increase the camber of the wing. Quote:
It shifts the whole polar to the right! That means it LOWERS our Angle of Attack!! You cannot have the same Angle of Attack flaps up as you do flaps down.... Check out figure 5: Quote:
It is not physically possible with TE flaps to have the same general CLmax presented by the RAE in clean configuration with a polar with the TE flaps down at the same Angle of Attack. Which incidentally also matches the 2D data from the NACA family of airfoils. BTW you can see the data point Mtt plotted for the plain airfoil and for the slats on the polar. The Bf-109 did not have full length LE slats so it did not get a dramatic CLmax increase. Last edited by Crumpp; 11-01-2011 at 10:15 PM. |
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#2
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#3
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In this case though you claimed that the top polars were different designs of TE flaps deployed and their effect. They were different designs of radiator flaps as I stated in my first reply of too many to you. Therefore, the curve in question on the bottom would be shifted to the right if that was the case. You started posting about the language used on the polar out of context and without the details. JtD your focus is never on the topic at hand. It is only to discredit anything I say in any way that you can. I have nothing further on this topic or any other topic for you. You can work whatever angle you dream up to claw at this conversation but I wish you good luck in your life. Quote:
It only becomes a problem when if the slats experience a mechanical malfunction and one slat cannot meet the force required. |
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#4
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I hve been thinking at the solution you wrote over the night and I hve some doubt of the solution proposed: V_SlatOut = VstallxSQR(G) At first I understand that this is similar to old IL2 and thus is a satisfactory solution for all. However my point here is that it cld be improved. Slat deployment on the 109 was governed by the air pressure on the leading edge (LE) and the hinged mechanism weight and frictions forces. a. Frictions forces are cte (K) b. Weight effect is dependent on G (P=mg) c. Dyn Pressue acting on the slat is a function of the speed of the plane (V) and the AoA (alpha) with Pdyn = 0.5roV²S*cos(alpha) Hence we have V_SlatOut = f(G, Pdyn) + K At 1G, the speed being known, as is the AoA we have the resulting value of the Weight and friction of the mechanism given that we make the calculation of the projected surface of the slat We can now choose to consider the friction of the mechanism negligible given tht the slat were known to be retractable only by the application of one finger (and much attention were required to keep the slat close on the ground to protect the mechanism from ingesting dust, sand and small objects). So basically we will hve V_SlatOut = f(G, Pdyn) tht result in the programmed law : If V<= V_Stall*SQR(G) and If Pdyn>=mg (m being the resulting balancing mass calculated at the 1G condition) Then Slats Out. The good thing is that by this way you hve an independent behaviour for both slat that can result in asymmetric deployment Pls note tht the Weight I am talking abt is not really a mass per G. It's the seen mass by the system combining all efforts in the mechanism that result in the deployment of the slat minus the friction. I am pulling away the frictions forces as they are not dependent of the G and are basically negligible if the system is functioning optimally. EDIT: Sry Crumpp I did delete my post as I needed to check my info. Here it is right as before. I checked the deployment principles here http://109lair.hobbyvista.com/index1024.htm |
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#5
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just a small vid i just made to spice this interesting thread up..
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#6
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Then why do you say the opposite, first?
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You could still explain how you see that the slats deploy within the 2.5° between 8 and 11.5° AoA as you said. From where I am standing, it is not on the chart. Last edited by JtD; 11-02-2011 at 05:22 PM. |
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#7
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Can you make one and stall the aircraft in 1G level flight and then stall it left/right in a level turn please? Also deploy the slats in turn by pulling the stick back and retract them by releasing it like I did on my aircraft to see what level of control you have. |
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#8
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My practical experience with 109 slats (actually, Buchon, so late version and no, I haven't flown them - lol) is that on the ground they're very definitely either up or down. The mechanism doesn't seem to hint at any kind of balance anywhere in between, it's all up or all down. They do seem rather secure in the up position while on ground but I could imagine with a plane bumping around on the grass during takeoff they would likely pop down, but there would be contributing air loads going on as well, obviously.
It also struck me while I was fiddling with them that it might take a little less AoA to pop them back in that what popped them out. If so, that would reduce the chance of having them banging back and forth if one happened to be riding the magic AoA. --- Let me know when I can groundloop a 109 (or anything else, for that matter - lol) in game. --- |
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#9
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That's the beauty of the design. They work automatically, allowing the pilot to maintain control at higher angles of attack without devoting mental effort to it.
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#10
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I understand here that friction wld be negligible as supposed. |
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